Date: Wed, 24 Sep 1997 11:13:27 -0600 (MDT) From: Nate Williams <nate@mt.sri.com> To: "Justin T. Gibbs" <gibbs@plutotech.com> Cc: Nate Williams <nate@mt.sri.com>, current@freebsd.org Subject: Re: new timeout routines Message-ID: <199709241713.LAA12839@rocky.mt.sri.com> In-Reply-To: <199709241709.LAA24417@pluto.plutotech.com> References: <199709241656.KAA12715@rocky.mt.sri.com> <199709241709.LAA24417@pluto.plutotech.com>
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> So you assume that regardless of what pointers the client gives you, > even if they give you the same pair twice without an intervening > expiration or untimeout call, that there will be no collisions in > the hash table? How did the original code in untimeout() determine what to pull off the table? Obviously there is enough information in the untimeout() call to uniquely determine which entry to use, and that same information was used in timeout(), so we must be able to build a perfect hash function. Nate
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